AgNO_{3} + KCl → AgCl + KNO_{3}

**$\mathbf{0}\mathbf{.}\mathbf{780}\mathbf{}\overline{)\mathbf{mg}\mathbf{}\mathbf{KCl}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{KCl}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mg}\mathbf{}\mathbf{KCl}}}\mathbf{\times}\frac{\mathbf{74}\mathbf{.}\mathbf{55}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{KCl}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{KCl}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{AgNO}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{KCl}}}$**

** = 0.05815 mol AgNO _{3}**

You may want to reference (Pages 149 - 153) Section 4.6 while completing this problem.

If 26.2 mL of AgNO_{3} is needed to precipitate all the Cl^{-} ions in a 0.780-mg sample of KCl (forming AgCl), what is the molarity of the AgNO_{3} solution?

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